You have found the following ages (in years) of all 4 lizards at your local zoo: $ 5,\enspace 1,\enspace 1,\enspace 3$ What is the average age of the lizards at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 lizards at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{5 + 1 + 1 + 3}{{4}} = {2.5\text{ years old}} $ Find the squared deviations from the mean for each lizard. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $5$ years $2.5$ years $6.25$ years $^2$ $1$ year $-1.5$ years $2.25$ years $^2$ $1$ year $-1.5$ years $2.25$ years $^2$ $3$ years $0.5$ years $0.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{6.25} + {2.25} + {2.25} + {0.25}} {{4}} $ $ {\sigma^2} = \dfrac{{11}}{{4}} = {2.75\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{2.75\text{ years}^2}} = {1.7\text{ years}} $ The average lizard at the zoo is 2.5 years old. There is a standard deviation of 1.7 years.